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Simplify Fractions again

Albert

Well-known member
Jan 25, 2013
1,225
$ a_n=(\dfrac{1}{\sqrt n+\sqrt {n-1}})\times(\dfrac{1}{\sqrt {n+1}+\sqrt {n-1}})\times(\dfrac{1}{\sqrt {n+1}+\sqrt n}) $
$S_n=a_1+a_2+a_3+-------+a_n$
$find:\,\, S_{2012}$
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Rationalizing the denominator of the expression \(\displaystyle a_n\) we get:

$ a_n=\left(\dfrac{1}{\sqrt n+\sqrt {n-1}}\cdot \dfrac {\sqrt n-\sqrt {n-1}}{\sqrt n-\sqrt {n-1}}\right)\times\left(\dfrac{1}{\sqrt {n+1}+\sqrt {n-1}}\cdot \dfrac {\sqrt {n+1}-\sqrt {n-1}}{\sqrt {n+1}-\sqrt {n-1}}\right)\times\left(\dfrac{1}{\sqrt {n+1}+\sqrt {n}}\cdot \dfrac {\sqrt {n+1}-\sqrt {n}}{\sqrt {n+1}-\sqrt {n}}\right) $

$ a_n=\left(\dfrac {\sqrt {n}-\sqrt {n-1}}{1}\right)\times \left(\dfrac {\sqrt {n}-\sqrt {n+1}}{-1}\right) \times \left(\dfrac {\sqrt {n+1}-\sqrt {n-1}}{2}\right)$


$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}\right)$

$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}-\sqrt{n}\right)+\dfrac{1}{2}\left(\sqrt{n}-\sqrt{n-1}\right)$

This is clearly a telescoping series and to compute \(\displaystyle S_{2012}\), we get:

\(\displaystyle S_{2012}=-\frac{1}{2}(\sqrt {2013}-\sqrt {1})+\frac{1}{2}(\sqrt{2012}-0)\)

\(\displaystyle S_{2012}=\frac{1}{2}-\frac{1}{2}(\sqrt{2013}-\sqrt{2012})\)
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
Rationalizing the denominator of the expression \(\displaystyle a_n\) we get:


$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}\right)$

$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}-\sqrt{n}\right)+\dfrac{1}{2}\left(\sqrt{n}-\sqrt{n-1}\right)$

This is clearly a telescoping series and to compute \(\displaystyle S_{2012}\), we get:

\(\displaystyle S_{2012}=-\frac{1}{2}(\sqrt {2013}-\sqrt {1})+\frac{1}{2}(\sqrt{2012}-0)\)

\(\displaystyle S_{2012}=\frac{1}{2}-(\sqrt{2013}-\sqrt{2012})-------(last \,\, step)\)
your last step is incorrect ,a typo happens
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
your last step is incorrect ,a typo happens
Yep, you're right Albert...I left off \(\displaystyle \frac{1}{2}\) in front of the surds, I'm sorry and I will fix my first post so that I get the correct answer to this problem.